" dx v(x) $ % You may also have seen this written in shorthand form as. & ' v(x) + u(x) '# % ! d

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1 Calculus II MAT 146 Mthods of Intgration: Intgration by Parts Just as th mthod of substitution is an intgration tchniqu that rvrss th drivativ procss calld th chain rul, Intgration by parts is a mthod of intgration that rvrss anothr drivativ procss, this on calld th product rul. Th Product Rul For two functions, ux) and vx), th product rul for drivativs stats that d dx [ ux)vx) ] =! # d dx ux) $! d ' vx) + ux) '# % dx vx) $. % You may also hav sn this writtn in shorthand form as du! v) = du! v + u! dv. In words, w might say that th drivativ of th product of two functions is th drivativ of th first function tims th scond function plus th first function tims th drivativ of th scond function. Whn w bgin with th shorthand rprsntation abov, w can crat a formula for intgration by parts by intgrating ach trm in th quation: du!v) = du! v + u!dv # du!v) = # du!v + # u!dv # duv) = # vdu + # udv uv = # vdu + # udv uv $ # vdu = # udv or # udv = uv $ # vdu Intgrat ach trm of th quation. Rwrit. On th lft sid, th intgral of a drivativ givs us th original product. Subtract vdu! from both sids of th quation. Switch lft-sid and right-sid xprssions traditional rprsntation). Th last quation in th abov drivation is th intgration-by-parts formula. It stats that th intgral of a product whos factors ar a function u and th drivativ of anothr function v is quivalnt to th diffrnc btwn th product of th two functions, u and v, and th intgral of th product composd of th function v and th drivativ of th function u.

2 Exampl 1: Us intgration by parts to valuat! x x dx, whru = x and dv = x dx. To apply th intgration-by-parts formula, w nd to dtrmin du and v. Th tabl format shown blow is a usful way to organiz th information: u = x dv = x dx du = dx v = x W startd th tabl by placing th known information, u and dv, in th top row. W compltd th tabl by calculating du and v and placing thm in th appropriat columns in th scond row. Whn th tabl is complt it bcoms a hlpful dvic for complting th intgration-by-parts task. u = x dv = x dx du = dx v = x From th tabl w writ th product of u and v ) and subtract from it th intgral of th product of v and du ). From th tabl, w gt! x x dx = x x! x dx W still ar lft with an intgral xprssion on th right sid of th quation, but this is on w can valuat by inspction:! x x dx = x x! x dx = x x x + C Exampl 2: Us intgration by parts to valuat! x ln x)dx. W hav two choics for th substitutions: u = x dv = lnx)dx u = ln x) dv = xdx du = dx v =? du = 1 x dx v = 1 2 x 2 Although our choic may not hav bn immdiatly clar at th start, by looking at th two options for substitutions, only on sms promising now. 1 W procd using th information in th right-hand tabl. 1 Th antidrivativ of lnx) dos xist. Drivd in Exampl 3, it is xlnx)-x. For most of us, howvr, that antidrivativ cannot b dtrmind by inspction, and that s a goal for intgration by parts.

3 u = ln x ) dv = xdx This lads to du = 1 x dx v = 1 2 x 2! x ln x)dx = ln x) 1 2 x 2 # $ 1 ' % 2 x2 ) $ 1 % x dx '! ) = 1 2 x 2 ln x) # 1! 2 xdx = 1 2 x 2 ln x ) # 1 2 $ 1 ' % 2 x2 ) + C = 1 2 x 2 ln x) # 1 4 x2 + C Exampl 3: Us intgration by parts to valuat! lnx)dx. Dtrmin th substitutions ndd for intgration by parts: u = ln x) dv = dx 1 This lads to du = 1 x dx v = x $ 1! lnx)dx = lnx) x # x % x dx '! ) 1 = x lnx) # dx! = x lnx) # x 1 = ln) # ) # 1ln1) #1) = # ) # 0 #1) = 1 In ths xampls, w s th importanc of our choics for u and dv. Thr ar thr important guidlins to kp in mind whn making ths choics. Whn choosing u, our goal should b an xprssion that simplifis whn it is diffrntiatd. If du is not a simplr xprssion than u, rconsidr your choic.

4 Your choic for dv should b an xprssion that can b asily intgratd s Exampl 2). Th intgral! vdu should b simplr to valuat that th original intgral! udv. Th nxt xampl illustrats that it could tak mor than on application of intgration by parts to valuat an intgral. Exampl 4: Evaluat! x sin xdx. Possibl substitutions: u = x dv = sin xdx u = sin x dv = x dx du = x dx v =! cos x du = cos xdx v = x Ths ar almost idntical, xcpt for a ngativ sign. W ll procd hr using th lft-hand tabl to illustrat th importanc of sign changs! x sin xdx = x cosx +! x cos xdx Th intgral on th right sid of th quation is similar to th on w startd with. It s valu is not radily apparnt, so w apply th intgration-by-parts tchniqu to it. For! x cos xdx, lt u = x and lt dv = cosxdx. Thn du = x dx and v = sinx. This givs us! x cos xdx = x sin x! x sin xdx. Substituting th right-hand xprssion for th lft in th first intgration-by-parts application, w gt! x sin xdx = x cosx +! x cos xdx ) = x cosx + x sin x x sin xdx! = x cosx + x sin x! x sin xdx W again sm to b lft with an intgral that rquirs furthr valuation. Notic, howvr, that th intgral on th lft sid of th quation is th sam as th on on th right sid. W can simplify th quation abov to complt th valuation of th original intgral:! x sin xdx = x cos x + x sin x! x sin xdx ) = x cos x + x sin x 2 x sin xdx! )! x sin xdx = 1 2 x cos x + x sin x = 1 2 x sin x cos x) + C

5 You might try valuating th original intgral of Exampl 4 using th altrnativ tabl of substitutions drivd at th bginning of th solution. What do you find? Is this what you xpctd?